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  3. 澳大利亚标准 AS3600 楼板设计实例及与 SkyCiv 的比较

澳大利亚标准 AS3600 楼板设计实例及与 SkyCiv 的比较

标准考虑的平板系统

澳大利亚标准确立了钢筋混凝土板设计的最低要求, 例如单向和双向类型. 关于平面配置和梁的包含, 楼板也可以分为四面支撑的楼板, 梁板系统, 平板, 和平板. 下图总结了这些类型.

AS3600 Slab Design Example

数字 1. 四面支撑板. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press).

AS3600 Slab Design Example

数字 2. 格栅板系统. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press).

AS3600 Slab Design Example

数字 3. 平板. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press).

AS3600 Slab Design Example

数字 4. 平板. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press).

标准推荐了一些方法 (简化且经过验证的程序) 在确定弯矩:

  • 条款 6.10.2: 连续梁和单向板
  • 条款 6.10.3: 四侧支撑的双向板
  • 条款 6.10.4: 具有多个跨度的双向板

该规范的目的是设计板系统中主要方向的钢筋总量. Rebar steel will be calculated for the bending moments “Mx” 和 “My.” 数字 5 shows the other forces or actions in a finite slab element in which the code prescribes their resistance values.

AS3600 Slab Design

数字 5. Forces in a finite slab element: bending moments (Mx, 我的), twisting moments (Mxy, Myx), and shears (Qx, Qy). (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

在这篇文章中, we will develop two slab design examples, one-way and two-way slab systems, using the simplified methods oriented and permitted by the code. In both instances, we will create a SkyCiv S3D model and compare the results against the methods mentioned above.

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One-Way Slab Design Example

Shown below is the small building and the slabs we will design

AS3600 One way slab design example

数字 6. One-way slabs in a small building example. (结构3D, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

AS3600 One way slab design example

数字 7. Plan dimensions and structural elements. (结构3D, SkyCiv Cloud Engineering).

For the slab example, in summary, the material, elements properties, and loads to consider :

  • Slab type classification: Oneway behaviour \(\压裂{L_2}{L_1} > 2 ; \压裂{14米}{6米}=2.33 > 2.00 \) 好的!
  • Building occupation: Residential use
  • Slab thickness \(t_{平板}=0.25m\)
  • Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \压裂{千牛}{m^3} + 0.6 \压裂{千牛}{m^3} \次 0.5 = 24.3 \压裂{千牛}{m^3} \)
  • Concrete characteristic compressive strength at 28 天 \(f'c = 25 兆帕 \)
  • Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 兆帕 \)
  • Slab Self-Weight \(Dead = \rho_w \times t_{平板} = 24.3 \压裂{千牛}{m^3} \times 0.25m = 6.075 \压裂 {千牛}{m^2}\)
  • Super-imposed dead load \(SD = 3.0 \压裂 {千牛}{m^2}\)
  • 活荷载 \(L = 2.0 \压裂 {千牛}{m^2}\)

Hand calculation according to AS3600 Standard

在这个部分, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strip.

  • Dead load, \(g = (3.0 + 6.075) \压裂{千牛}{m^2} \次 1 m = 9.075 \压裂{千牛}{米}\)
  • 活荷载, \(q = (2.0) \压裂{千牛}{m^2} \次 1 m = 2.0 \压裂{千牛}{米}\)
  • Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\次 9.075 + 1.5\次 2.0)\压裂{千牛}{米} =13.89 \frac{千牛}{米} \)

Using the simplified method specified by the standard, 第一, it is a must to comply with the following restrictions:

  • \(\压裂{L_i}{L_j} \这 1.2 . \压裂{6米}{6米} =1 < 1.2 \). 好的!
  • Load has to be uniform. 好的!
  • \(q \le 2g. q=2 \frac{千牛}{米} < 18.15 \压裂{千牛}{米}\). 好的!
  • The slab cross-section has to be uniform. 好的!.

Recommended minimum thickness, d

\(d \ge \frac{L_{fe}}{{k_3}{k_4}{\sqrt[3]{\压裂{\压裂{\三角洲}{L_{ef}}{E_c}}{F_{d, ef}}}}}\)

在哪里

  • \(k_3 = 1.0; k_4 = 1.75 \)
  • \(\压裂{\三角洲}{L_{ef}}=1/250 \)
  • \(E_c = 27600 兆帕 \)
  • \(F_{d,ef} = (1.0 +钢底板设计欧洲规范{cs})\times g + (\psi_s + 钢底板设计欧洲规范{cs}\times \psi_1) \times q=(1.0+0.8)\次 9.075 + (0.7+0.8\次 0.4)\次 2 = 18.375 kPa\)
    • \(\psi_s = 0.7 \) Live-load short-term factor
    • \(\psi_1 = 0.4 \) Live-load long-term factor
    • \(钢底板设计欧洲规范{cs} = 0.8 \)

\(d \ge \frac{5.50米}{{1.0}\次 {1.75}{\sqrt[3]{\压裂{\压裂{1}{250}\次{27600 \times 10^3 kPa}}{18.375 千帕}}}} \ge 0.173m. d = 0.25m > 0.173米 \) 好的!

Once we demonstrate that constraints are satisfied, the bending moment is calculated using the expression: \(M=\alpha \times F_d \times L_n^2\) 哪里 \(\alpha\) is a constant defined in the following figure.

 

bending moment coefficients for one way slab design according AS3600

数字 8. Values of moment coefficient \(\alpha\) for slabs with more than two spans. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press).

在哪里:

  • (一个) Case of slabs and beams on girder support
  • (b) For continuous beam support only
  • (C) Where Class L reinforcement is used
  • \(L_n \) is the unitary strip span
  • \(F_d \) is the gravitational factored load

For the slab example, we have to use case (一个) because the slab rests on stiff girders. It will be explained only one case and the rest will show in the following table. We include also the steel reinforcement area calculation.

  • \(M={\α} {F_d}{L_n^2}={-\压裂{1}{24}}\次 {13.89 \压裂{千牛}{米}}\次 (6米-0.5米)可以假设为 – 17.51{千牛}{米}\)
  • Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 分钟).
  • \(d = t_{平板} – 覆盖 – \压裂{BarDiameter}{2} = 250mm – 20毫米 – 6mm = 224mm \)
  • \(\alpha_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \le \alpha_2 \le 0.85) \) 从而, we select \(\alpha_2 = 0.85\)
  • \(\xi = \frac{\alpha_2\times f’c}{F_{他的}} = frac{0.85\次 25 兆帕}{500 兆帕} = 0.0425 \)
  • \(\rho_t = \xi – \sqrt{{\习}^ 2 – \压裂{{2}{\习}{中号}}{{\φ}{b}{d^2}{F_{他的}}}} = 0.0425 – \sqrt{{0.0425}^2-\frac{2\times 0.0425\times 17.51{千牛}{米}}{{0.8}\次 {1米}\次 {{(0.224米)^ 2}} \次 {500\次 {10^ 3}千帕}}}=0.0008814\)
  • \(\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le \gamma \le 0.85) \) 从而, we select \(\gamma = 0.85\)
  • \(k_u = \frac{\rho_t \times f_{他的}}{0.85\times \gamma \times f’c}= 分数{0.0008814\次 500 兆帕}{0.85\次 0.85 \次 25 兆帕} =0.0244\)
  • \(\phi = 1.19 – \压裂{13\钢底板设计欧洲规范{u0}}{12} = 1.19 – \压裂{13\次 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8) \) 从而, we select \(\phi = 0.8\). 好的!.
  • \(\o{Ť,分} = 0.20 {(\压裂{d}{d})^ 2}{(\压裂{F'_{ct,F}}{F_{他的}})} = 0.20 \次 (\压裂{0.25米}{0.224米})^2 \times \frac{0.6\次 sqrt{25兆帕}}{500 兆帕} = 0.0015\)
  • \(一个_{圣}=max(\o{Ť,分}, \rho_t)\times b \times d = max(0.0015,0.0008814)\次 1000 mm \times 224 mm = 334.82 mm^2 \)
\(\alpha\) and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
\(\alpha\) 值 -\(\压裂{1}{24}\) \(\压裂{1}{11}\) -\(\压裂{1}{10}\) \(\压裂{1}{10}\) \(\压裂{1}{16}\) \(\压裂{1}{11}\)
M value -17.51 38.20 -42.02 42.02 26.26 38.20
\(\rho_t\) 0.0008814 0.001948 0.002148 0.002148 0.00133 0.001948
ku 0.0244 0.0539 0.0594 0.0594 0.0368 0.05391
\(\phi\) 0.8 0.8 0.8 0.8 0.8 0.8
\(一个_{圣} {mm^2}\) 334.82 436.31 481.099 481.099 334.8214 436.3100

After the steel rebar area calculation, you can define the detailing (the actual way to place the reinforcement into the slab). As help for your knowing, we share the following image, which indicates the rebar location for positive and negative moments:

AS3600 slab design example

数字 9. Reinforcement arrengement for one-way and two-way slabs. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

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SkyCiv S3D Plate Design Module Results

In the first view, we will show some images for the modeling and structural analysis of the example in S3D. We recommend you read about modeling in SkyCiv in the following links How to model plates?ACI Slab Design Example with SkyCiv.

AS3600 One way slab design model

数字 10. Structural Model in S3D for one-way slabs example. (结构3D, SkyCiv Cloud Engineering).

Before analyzing the model, we must define a plate mesh size. Some references (2) recommend a size for the shell element of 1/6 of the short span or 1/8 of the long span, the shorter of them. Following this value, 我们有 \(\压裂{L2}{6}= 分数{6米}{6} = 1m \) 要么 \(\压裂{L1}{8}= 分数{14米}{8}=1.75m \); we take 1m as a maximum recommended size and 0.50m applied mesh size.

AS3600 One way slab design model

数字 11. Improved mesh in plates. (结构3D, SkyCiv Cloud Engineering).

Once we improved our analytical structural model, we run a linear elastic analysis. When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of \(\压裂{大号}{250}= 分数{6000毫米}{250}=24.0 mm\).

AS3600 One way slab design model

数字 12. Vertical displacement in plates. (结构3D, SkyCiv Cloud Engineering).

Comparing the maximimum vertical displacement against the code referenced value, the slab’s stiffness is adequate. \(4.822 毫米 < 24.00mm\).

The maximum moments in the slab’s spans are located for positive in the center and for negative at the exterior and interior supports. Let’s see these moments values in the following images.

AS3600 One way slab design model

数字 13. Moments in the X direction. (结构3D, SkyCiv Cloud Engineering).

AS3600 One way slab design model

数字 14. Moments in the Y direction. (结构3D, SkyCiv Cloud Engineering).

Plate element local axes are indicated below.

AS3600 One way slab design model

数字 15. Slab local axes. (结构3D, SkyCiv Cloud Engineering).

For more details about automated reinforced slab design, see our documentation Plates in SkyCiv.

AS3600 slab design

数字 16. Top D1 reinforcement. (结构3D, SkyCiv Cloud Engineering).

AS3600 slab design

数字 17. Bottom D1 reinforcement. (结构3D, SkyCiv Cloud Engineering).

AS3600 slab design

数字 18. Top D2 reinforcement. (结构3D, SkyCiv Cloud Engineering).

AS3600 slab design

数字 19. Bottom D2 reinforcement. (结构3D, SkyCiv Cloud Engineering).

Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis (local axes “2”) and handcalculations.

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
\(一个_{圣, HandCalcs} {mm^2}\) 334.82 436.31 481.099 481.099 334.8214 436.3100
\(一个_{圣, S3D} {mm^2}\) 285.13 313.00 427.69 427.69 313.00 427.69
\(\三角洲_{dif}\) (%) 14.84 28.262 11.101 11.101 6.517 1.986

We can see that the results of the values are very close to each other. This means the calculations are correct!

Two-way Slab Design Example

在这个部分, we will develop an example that consists of a grillage system.

AS3600 two way slab design example

数字 20. Grillage System. (结构3D, SkyCiv Cloud Engineering).

The plan dimensions are shown at next

AS3600 two way slab design example

数字 21. Plan dimensions for the four sides two-way slab example. (结构3D, SkyCiv Cloud Engineering).

For the slab example, in summary, the material, elements properties, and loads to consider :

  • Slab type classification: Twoway behaviour \(\压裂{L_2}{L_1} \这 2 ; \压裂{7米}{6米}=1.167 < 2.00 \) 好的!
  • Building occupation: Residential use
  • Slab thickness \(t_{平板}=0.25m\)
  • Reinforced concrete density assuming a steel reinforcement ratio of 0.5% \(\rho_w = 24 \压裂{千牛}{m^3} + 0.6 \压裂{千牛}{m^3} \次 0.5 = 24.3 \压裂{千牛}{m^3} \)
  • Concrete characteristic compressive strength at 28 天 \(f'c = 25 兆帕 \)
  • Concrete Modulus of Elasticity by Australian Standard \(E_c = 26700 兆帕 \)
  • Slab Self-Weight \(Dead = \rho_w \times t_{平板} = 24.3 \压裂{千牛}{m^3} \times 0.25m = 6.075 \压裂 {千牛}{m^2}\)
  • Super-imposed dead load \(SD = 3.0 \压裂 {千牛}{m^2}\)
  • 活荷载 \(L = 2.0 \压裂 {千牛}{m^2}\)

Hand calculation according to AS3600 Standard

在这个部分, we will calculate the required reinforced steel rebar using the reference of the Australian Standard. We first obtain the total factored bending moment to be carried out by the slab’s unitary width strips in each bending main direction.

  • Dead load, \(g = (3.0 + 6.075) \压裂{千牛}{m^2} \次 1 m = 9.075 \压裂{千牛}{米}\)
  • 活荷载, \(q = (2.0) \压裂{千牛}{m^2} \次 1 m = 2.0 \压裂{千牛}{米}\)
  • Ultimate load, \(Fd = 1.2\times g + 1.5\times q = (1.2\次 9.075 + 1.5\次 2.0)\压裂{千牛}{米} =13.89 \frac{千牛}{米} \)

Design moments and coefficients

AS3600 two way slab design example

数字 22. Orientation of a two-way slab for positive moments determination. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

AS3600 two way slab design example

数字 23. Negative moments determination in a two-way slab. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

Edge Condition Short-span coefficients (\(\beta_x\)) Long-span coefficients (\(\beta_y)\) all values of \(\压裂{L_y}{L_x}\)
的价值 \(\压裂{L_y}{L_x}\)
1.0 1.1 1.2 1.3 1.4 1.5 1.75 \(\葛 2.0\)
1. Four edges continuous 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024
2. One short edge discontinuos 0.028 0.032 0.036 0.038 0.041 0.043 0.047 0.050 0.028
3. One long edge discontinous 0.028 0.035 0.041 0.046 0.050 0.054 0.061 0.066 0.028
4. Two short edges discontinous 0.034 0.038 0.040 0.043 0.045 0.047 0.050 0.053 0.034
5. Two long edges discontinous 0.034 0.046 0.056 0.065 0.072 0.078 0.091 0.100 0.034
6. Two adjacent edges discontinous 0.035 0.041 0.046 0.051 0.055 0.058 0.065 0.070 0.035
7. Three edges discontinuous (one long edge continuous) 0.043 0.049 0.053 0.057 0.061 0.064 0.069 0.074 0.043
8. Three edges discontinuous (one short edge continous) 0.043 0.054 0.064 0.072 0.078 0.084 0.096 0.105 0.043
9. Four edges discontinuos 0.056 0.066 0.074 0.081 0.087 0.093 0.103 0.111 0.056

桌子 1. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

The following image explain the all nine cases that the table above refers

AS3600 two way slab design example

数字 24. Edge conditions for two-way slabs supported on four sides. (红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press)

Design moments for central region (案件 6 Two adjacent edges discontinuous) :

  • \(L_x = 6m, L_y=7m, \压裂{L_y}{L_x} = frac{7米}{6米}= 1.167 \) Values to be linearly interpolated
  • Positives:
    • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.04435}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}=22.177 kNm\)
    • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.035}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}=17.501 kNm \)
  • Negatives exterior span:
    • \(M_{x1,A} = -\lambda_e \times M_x = -0.5 \次 22.177 kNm = – 11.089 kNm\)
    • \(M_{y1,A} = -\lambda_e \times M_y = -0.5 \次 17.501 kNm = -8.751 千牛·米 \)
  • Negatives interior span:
    • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \次 22.177 kNm = – 29.495 kNm\)
    • \(M_{y1, 乙} = -\lambda_{1和} \times M_y = -1.33 \次 17.501 kNm = -23.276 千牛·米 \)

Design moments for central region (案件 3 One long edge discontinous) :

  • \(L_x = 6m, L_y=7m, \压裂{L_y}{L_x} = frac{7米}{6米}= 1.167 \) Values to be linearly interpolated
  • Positives:
    • \(M_x = {\beta_x}{F_d}{L_x^2} = {0.03902}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}= 19.512 kNm\)
    • \(M_y = {\beta_y}{F_d}{L_x^2} ={0.028}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}= 14.001 千牛·米 \)
  • Negatives interior span:
    • \(M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \次 19.512 kNm = – 25.951 kNm\)
    • \(M_{y1,B} = -\lambda_{1和} \times M_y = -1.33 \次 14.001 kNm = – 18.621 千牛·米 \)
  • Negatives interior second span:
    • \(M_{x2,B} = -\lambda_{2X} \times M_x = -1.33 \次 19.512 kNm = – 25.951 kNm\)
    • \(M_{y2,B} = -\lambda_{2和} \times M_y = -1.33 \次 14.001 kNm = – 18.621 千牛·米 \)

Rebar steel for X direction

\(\alpha\) and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 11.089 22.177 29.495 25.951 19.512 25.951
\(\rho_t\) 0.00055614 0.00112 0.001496 0.001313 0.000984 0.001313
ku 0.015395 0.0310 0.0414 0.0364 0.0272 0.0364
\(\phi\) 0.8 0.8 0.8 0.8 0.8 0.8
\(一个_{圣} {mm^2}\) 334.8214 334.8214 335.08233 334.821 334.8214 334.8214

Rebar steel for Y direction

\(\alpha\) and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 8.751 17.501 23.276 18.621 14.001 18.621
\(\rho_t\) 0.0004383 0.0008811 0.001176 0.0009381 0.000703 0.0009381
ku 0.0121 0.0244 0.03256 0.02597 0.0195 0.02597
\(\phi\) 0.8 0.8 0.8 0.8 0.8 0.8
\(一个_{圣} {mm^2}\) 334.821 334.821 334.821 334.821 334.8214 334.821

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SkyCiv S3D Plate Design Module Results

After refining the model, is time to run a linear elastic analysis.

When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of \(\压裂{大号}{250}= 分数{6000毫米}{250}=24.0 mm\).

SkyCiv S3D Plate Design Module Results

数字 25. Vertical Displacement in the grillage slab system. (结构3D, SkyCiv Cloud Engineering).

The image above gaves to us the vertical displacement. The maximum value is -1.179mm being less than the maximum allowed of -24mm. 因此, the slab’s stiffeness is adequate.

SkyCiv S3D Plate Design Module Results

数字 26. Plates moments in the X direction. (结构3D, SkyCiv Cloud Engineering).

Images 27 和 28 consist of the bending moment in each main direction. Taking the moment distribution and values, the software, SkyCiv, can obtain then the total steel reinforcement area.

SkyCiv S3D Plate Design Module Results

数字 27. Plates moments in the Y direction. (结构3D, SkyCiv Cloud Engineering).

Steel reinforcement areas:

SkyCiv S3D Plate Design Module Results

数字 28. Top Steel Rebar Reinforcement in Direction 1. (结构3D, SkyCiv Cloud Engineering).

SkyCiv S3D Plate Design Module Results

数字 29. Bottom Steel Rebar Reinforcement in Direction 1. (结构3D, SkyCiv Cloud Engineering).

SkyCiv S3D Plate Design Module Results

数字 30. Top Steel Rebar Reinforcement in Direction 2. (结构3D, SkyCiv Cloud Engineering).

SkyCiv S3D Plate Design Module Results

数字 31. Bottom Steel Rebar Reinforcement in Direction 2. (结构3D, SkyCiv Cloud Engineering).

Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis and handcalculations.

Rebar steel for X direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
\(一个_{圣, HandCalcs} {mm^2}\) 334.8214 334.8214 335.08233 334.821 334.8214 334.8214
\(一个_{圣, S3D} {mm^2}\) 289.75 149.35 325.967 325.967 116.16 217.311
\(\三角洲_{dif}\) (%) 13.461 55.39 2.720 2.644 65.307 35.0964

Rebar steel for Y direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
\(一个_{圣, HandCalcs} {mm^2}\) 334.821 334.821 334.821 334.821 334.821 334.821
\(一个_{圣, S3D} {mm^2}\) 270.524 156.75 304.34 304.34 156.75 270.52
\(\三角洲_{dif}\) (%) 19.203 53.184 9.104 9.104 53.184 19.204

The diference is some high for positive moments and the reason would be the presence of beams with high torsional stiffness that impact on the Plate Finite Element Analysis Results and the calculations for bending reinforcement steel.

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参考资料

  1. 红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press.
  2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1编辑, LIMUSA.
  3. 澳大利亚标准, 混凝土结构, 如 3600:2018

 

 

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